advanced-calculus-evaluate-0-x-2-e-x-2-ln-x-

Question Number 137048 by mnjuly1970 last updated on 29/Mar/21

\dots \dots . . a d v a n c e d \dots \dots \dots c a l c u l u s \dots \dots .

e v a l u a t e ::::

\boldsymbol χ = \int_{0}^{\infty} x^{2} e^{- x^{2}} l n (x) = ? ? ?

Answered by Ar Brandon last updated on 29/Mar/21

\boldsymbol χ = \int_{0}^{\infty} x^{2} e^{- x^{2}} lnxdx = \frac{\partial}{\partial t} ∣_{t = 0} \int_{0}^{\infty} x^{2 + t} e^{- x^{2}} dx

\overset{(u = x^{2})}{=} \frac{\partial}{\partial t} ∣_{t = 0} \frac{1}{2} \int_{0}^{\infty} u^{\frac{1}{2} + \frac{t}{2}} e^{- u} du = \frac{\partial}{\partial t} ∣_{t = 0} \frac{1}{2} Γ (\frac{3}{2} + \frac{t}{2})

= \frac{1}{2} ∣_{t = 0} \frac{1}{2} Γ (\frac{3}{2} + \frac{t}{2}) ψ (\frac{3}{2} + \frac{t}{2})

= \frac{1}{4} Γ (\frac{3}{2}) ψ (\frac{3}{2}) = \frac{1}{4} \cdot \frac{\sqrt{π}}{2} \cdot (2 - γ - 2 \ln 2)

Commented by mnjuly1970 last updated on 29/Mar/21

v e r y v e r y n i c e

m r b r a n d o n . . t h a n k y o u \dots

Commented by Ar Brandon last updated on 29/Mar/21

My pleasure, Sir

Answered by Dwaipayan Shikari last updated on 29/Mar/21

\int_{0}^{\infty} x^{2} e^{- x^{2}} l o g (x) d x = ℵ^{'} (3)

ℵ (α) = \int_{0}^{\infty} x^{α - 1} e^{- x^{2}} d x \Rightarrow ℵ (α) = \frac{1}{2} \int_{0}^{\infty} j^{\frac{α}{2} - 1} e^{- j} d j

ℵ (α) = \frac{Γ (\frac{α}{2})}{2} \Rightarrow ℵ^{'} (α) = \frac{Γ^{'} (\frac{α}{2})}{4} \Rightarrow ℵ^{'} (3) = \frac{\sqrt{π}}{8} (- γ + 2 - 2 l o g (2))

Commented by mnjuly1970 last updated on 29/Mar/21

t h a n k s a l o t m r p a y a n \dots

Commented by Dwaipayan Shikari last updated on 29/Mar/21

Have a great day sir!

Answered by mathmax by abdo last updated on 30/Mar/21

Φ = \int_{0}^{\infty} x^{2} e^{- x^{2}} lnx dx let f (a) = \int_{0}^{\infty} x^{a} e^{- x^{2}} dx

v (o) x^{a} e^{- x^{2}} \sim x^{a} \int_{0}^{ξ} \frac{dx}{x^{- a}} cv \Leftrightarrow - a < 1 \Rightarrow a > - 1

we have f (a) = \int_{0}^{\infty} e^{alogx} e^{- x^{2}} dx \Rightarrow f^{^{'}} (a) = \int_{0}^{\infty} x^{a} e^{- x^{2}} logx dx

and f^{^{'}} (2) = \int_{0}^{\infty} x^{2} e^{- x^{2}} logx dx

changement x^{2} = t givef (a) = \int_{0}^{\infty} t^{\frac{a}{2}} e^{- t} \frac{dt}{2 \sqrt{t}}

= \frac{1}{2} \int_{0}^{\infty} t^{\frac{a - 1}{2}} e^{- t} dt = \frac{1}{2} \int_{0}^{\infty} t^{\frac{a - 1}{2} + 1 - 1} e^{- t} dt

= \frac{1}{2} \int_{0}^{\infty} t^{\frac{a + 1}{2} - 1} e^{- t} dt = \frac{1}{2} Γ (\frac{a + 1}{2}) \Rightarrow f^{^{'}} (a) = \frac{1}{4} Γ^{^{'}} (\frac{a + 1}{2}) \Rightarrow

f^{^{'}} (2) = \frac{1}{4} Γ^{^{'}} (\frac{3}{2}) = Φ